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If light of frequency 8.2xx10^14 Hz is i...

If light of frequency `8.2xx10^14` Hz is incident on the metal , cut-off voltage for photoelectric emission is 2.03 V.Find the threshold frequency . Given h=`6.63xx10^(-34) ` Js , e=`1.6xx10^(-19)` C.

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To find the threshold frequency (\( \nu_0 \)) for the photoelectric emission from the given data, we can use the photoelectric equation: \[ E V = h \nu - h \nu_0 \] Where: - \( E \) is the charge of an electron (\( 1.6 \times 10^{-19} \) C), ...
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